Q1. 1> An aircraft takesoff from A (30N 020 W) maintains the great circle track to B (40N 020E). Given Rhum line track A to B is 73 degrees.find the great circle track at B.
84 28 T – Route A to B is Eastbound in Northern Hemisphere. Mean Lat = 35N. dLong = 40. Conversion Angle (CA) = 0.5 * 40 * sin(35) ≈ 11.47°. For NH Eastbound, Final GC Track > RL Track. Track B = 73 11.47 = 84.47° ≈ 84°28′.
Q2. 2> Given Rhum line track A (40N 020 W) to B (50N 060W) is 290 degrees. Aircraft takes off from A and follows the great circle. Find the great circle track at B.
275 52 T – Route A to B is Westbound in Northern Hemisphere. Mean Lat = 45N. dLong = 40. CA = 0.5 * 40 * sin(45) ≈ 14.14°. For NH Westbound, Final GC Track < RL Track. Track B = 290 – 14.14 = 275.86° ≈ 275°52′.
Q3. 3> Given Rhum line track A (60S 120 W) to B (50S 060W) is 74 degrees. Aircraft takeoff from A and follows the great circle. find the great circle track at A.
98 34 T – Route A to B is Eastbound in Southern Hemisphere. Mean Lat = 55S. dLong = 60. CA = 0.5 * 60 * sin(55) ≈ 24.57°. For SH Eastbound, Initial GC Track > RL Track. Track A = 74 24.57 = 98.57° ≈ 98°34′.
Q4. 4> Aircraft takeoff from A and follows the great circle track to B. find the great circle track at A. Given Rhum line track B (30 S 020 W) to A (40S 060E) is 99 degrees.
256 04 T – First, find RL A to B (Westbound). RL B->A is 099, so RL A->B is 279. Mean Lat = 35S. dLong = 80. CA = 0.5 * 80 * sin(35) ≈ 22.94°. SH Westbound: Initial GC < RL. Track A = 279 – 22.94 = 256.06° ≈ 256°04′.
Q5. 5> An aircraft autopilot is engaged to the route A (40 N 170 W) to B (30N 160W). Find a great circle track at when aircraft reaches B. Given Rhum line track B to A is 321 degrees.
143 52 T – First, find RL A to B (Eastbound). RL B->A is 321, so RL A->B is 141. Mean Lat = 35N. dLong = 10. CA = 0.5 * 10 * sin(35) ≈ 2.87°. NH Eastbound: Final GC > RL. Track B = 141 2.87 = 143.87° ≈ 143°52′.
Q6. 6> Given Rhum line track B (40S 000W) to A (60S 120E) is 105 degrees. Find the initial great circle for route B to A.
150 57 T – Route B to A is Eastbound in SH. Mean Lat = 50S. dLong = 120. CA = 0.5 * 120 * sin(50) ≈ 45.96°. SH Eastbound: Initial GC > RL. Track B = 105 45.96 = 150.96° ≈ 150°57′.